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Chemical Structure and Reactivity

by: Selena Trantow

Chemical Structure and Reactivity CHEM 3A

Marketplace > University of California - Berkeley > Chemistry > CHEM 3A > Chemical Structure and Reactivity
Selena Trantow

GPA 3.65

S. Pedersen

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S. Pedersen
Class Notes
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This 8 page Class Notes was uploaded by Selena Trantow on Thursday October 22, 2015. The Class Notes belongs to CHEM 3A at University of California - Berkeley taught by S. Pedersen in Fall. Since its upload, it has received 57 views. For similar materials see /class/226747/chem-3a-university-of-california-berkeley in Chemistry at University of California - Berkeley.


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Date Created: 10/22/15
Midterm 2 Study Guide This does not cover everything but it covers most important concepts Radical Halogenation always has three main phases hv light CH4 Br2 gt CHsBr HBr Initiation start of the reaction form 2 radicals from no radicals Break a weak bond eg OO bond 94 gt 2 Br39 Propogation the chainreaction part of the reaction must regenerate the radical you start with start with Br39 end with Br39 H3cquott r a H3CBr B Termination the end of the reaction form no radicals from 2 radicals Some Possibilities H3C39 j CH3 gt H3CCH3 start with 2 radicals f end with no radicals 39 quot H3C39 jBr gt H3CBr A A Br Br Br2 Selectivity when there are different kinds of H39s on the molecule selectivity trend 3 H gt 2 H gt 1 H Br2 hv A Br Br major product Always bear in mind that if you make a stereocenter consider all possible stereoisomers A Biz hV Br Br CI F 3 CI F CIAF Stereochemistry definitions to know chiral a moleculeobject whose mirror image is nonsuperimposable eg your hands stereoisomers molecules w identical connectivities different orientations in space Very general term enantiomers molecules that are nonsuperimposable mirror images of each other F pl X MeHEHMe diastereomers molecules that are stereoisomers but not enantiomers require 2 stereocenters CH3 939 9H3 9 3 7 7 2 stereocenters are the same 1 is flipped NH2 NH2 stereocenter an atom with four different groups around it H stereocenters are R or S Br CICH3 meso a molecule w stereocenters but which is achiral has an internal mirror plane Br Br Br 39I If you swap ANY two groups on the stereocenter you make the opposite stereocenter If you move three groups it39s the same as swapping two and then swapping two you get back the original stereocenter Miscelleneous Tips The best way to determine if molecules are enantiomers diasteromers or neither is to compare stereocenters directly if all stereocenters are identical the molecules are identical if all stereocenters are flipped the molecules are enantiomers if some stereocenters are flipped and some are identical the molecules are diastereomers this does not include molecules that may be identical because they are hidden rotations of each other etc From these rules it follows that if a molecule has only 1 stereocenter it has an enantiomer but no diastereomers Example CH3 939 9H3 939 A B c lZlZSEZEEZFSEEIEZEESJli ISrETlSEQSIQlimi ii 39z NH2 NH2 stereocenter C is identical from molecule 1 to molecule 2 molecule 1 molecule 2 Molecules 1 2 are DIASTEREOMERS Assigning stereocenters groups get priority by molecular weight Put 4th priority group pointing back k lf l to 3 goes clockwise R m Br CICH3 E Br CH3 lf1 to 3 goes COW S This stereocenter is R E like a Newman prof Fisher Projections B B B D A C E Ago E v D 5 Axe Things to keep in mind enantiomers 90 degree rotation of Fisher projection makes the ENANTIOMER A D B 08 E 08 E x c c A C 180 degree rotation of Fisher projection makes the ORIGINAL MOLECULE equivalent to two 90 degree rotations first makes the enantiomer second makes the enantiomer again which is the first molecule again Some other facts to consider swapping three groups on the stereocenter makes the same stereocenter again these don39t seem identical but they are because the top stereocenter is identical in both cases Just moved 3 groups around these don39t seem identical but they are because the swapped OH heads at topbottom aren39t stereocentersl They can go anywhere Substitution Reactions Important tidbits about the gig Reaction Substitution Nucleophilic Bimolecular General mechanism is concerted with backside attack and inversion of stereochernistry A A Nuce Egg e NUC quotB rate kNu E 0 Transition state is pentacoordinate with the electrophilic center flattened sp2like A Nuc LG Order of reactivity of electrophiles toward 8N2 conditions IMPORTANT GENERAL TREND 03 O 1 electrophiles gt 2 electrophiles gt 3 electrophiles one important exception triple beta branching is really bad could break this trend The above trend is due to sterics More steric congestion slows the reaction Prefers strong nucleophiles and a polar aprotic solvent Important tidbits about the s31 Reaction Substitution Nucleophilic Unimolecular General mechanism is two steps with carbocation formation and mostly racemization of stereocenter e F Nuc gt Bu gt vaG CA B Nuc Nuc B C C C eNuc 5050 of retentioninversion rate kE Order of reactivity of electrophiles toward SN1 conditions IMPORTANT GENERAL TREND l 3 electrophiles gt 2 electrophiles gt 1 electrophiles 3 carbocations gt 2 carbocations gt 1 carbocations The above trend is due to carbocation stabilization More stable carbocation forms faster Prefers weak nucleophiles and a polar protic solvent andor ionizing conditions eg Ag Elimination Reactions These reactions happen alongside the substitution reactions E1 and SN1 will compete E2 and SN2 will compete E2 Reaction Elimination Bimolecular Happens alongside the SN2 reaction General Mechanism HOe H m concerted gt S Br gt antiperiplanar The reaction is concerted and has a strict geometric requirement of antiperiplanar synperiplanar can happen when there39s no other choice but it absolutely must always be periplanar Q Br H Br Br H 1 l v I 63 I CD BA D AB CD AB H AB The stereochemistry of the product is this conformer this conformer this conformer readily will not do E2 would only slowly does E2 and is the determined by the do E2 and in fact only conformer to react conformation that the d es n t39 E2 happens in E1 Reaction Elimination Unimolecular Happens alongside the SN1 reaction General Mechanism Can make lots of different EtOH 9 heat QJQ 39 A elimination products depending on how many different H39s are j elimination product on adjacent carbons EtoH What some general trends things t knowto predict SN1I SN2I g E2 More substituted electrophiles prefer SN1 over SN2 3 electrophiles NEVER do SN2 Less substituted electrophiles prefer SN2 over SN1 1 electrophiles NEVER do SN1 Similarly more substituted electrophiles prefer E1 over E2 Less substituted electrophiles prefer E2 over E1 Strong bases prefer SN2 or elimination E1 or E2 SN1 will NEVER take place with a strong base Strong bases will start to prefer elimination over SN2 as the electrophile gets more hindered Thus a very unhinderednot substituted electrophile strong base will give SN2 a very hinderedsubstituted electrophile strong base will give elimination Weak bases prefer SN1 and generally disprefer elimination Strongly hindered bases will always prefer elimination over substitution At the end of the day you39re choosing two things is it favorable to make a carbocation gt if so E lSN l will be favored if not SN2E2 will be favored is there too much steric hindrance to do substitution gt if so elimination is favored Carbocation Shifts They happen Be aware You can have a hydride shift or an alkyl shift and they only happen for a good reason eg formation of a more stable carbocation Be sure to count all atoms in the area before and after doing a shift to make sure you didn39t mess up Example x L J What39s the mechanism AgNO3 HO CI OH2 H HJY XV JV CI H e H H H H199 a 2 cation a 3 cation JH 2 chloride ambiguous more stable H20 BUT weak nucleophile AgNO3 8N1 conditions Tips for Mechanisms Ultimately solving any mechanism requires a flash of insight a point where you just see the critical step that gets you from the starting material to the product However there are a few things you can do to maximize your chances of having the insight Atom Mapping Figure out as best you can which carbonsatoms in the sm are which carbonsatoms in the products Don39t overdo it but try to pick out some obvious ones Carbon numbering Often the backbone of the molecule doesn39t change much from starting material to product Other times it does change but in a way you can figure out If you can figure out how the backbone maps from sm to product number the carbons in both so you can more easily see exactly where changes are happening Try to think about the changes in terms of chemistry you39ve already learned For example if you notice a methyl group has moved from the sm to the product consider that you might have a carbocation rearrangement in the form of a methy lshift PLAY AROUND This is how we first figure out mechanisms in the real world anyway we draw some structures on the board try to envision some reasonablelooking reactions happening and once we have an answer we run experiments in the lab to support or disprove the mechanism If you think something might happen and it seems reasonable just draw it see if it seems to lead you on a good path or a dead end Problems To Be Worked On The Board 1 Predict the products of the following reations Where relevant show all stereoisomers Pay particular attention to any instructions given in the product boxes Pg 136 2nd problem Br MeOH heat Pg 177 last problem 2 For the molecule below how man sereocenters are there Circle each one J Based on the number of stereocenters circled what is the theoretical max number of stereoisomers possible K For this particular molecule it is IMPOSSIBLE to have the maximum number of stereoisomers calculated in Part J Explain L For the molecule shown below draw the enantiomer and one diastereomer in the boxes provided Skeletons are provided for you in the boxes to make it easier all possible products including stereoisomers with the molecular formula 09H17CI pg 249 L O O enantiomer O O diastereomer 3 Write a logical arrowpushing mechanism for the following reactions Br heat HINT the theme of this problem was oarbocation chemistry Pg 35 2nd problem Br heat W I MeOH O 9 HO o s CH3 HO CH3 395 0 Pg 21 2nd problem


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